To find the surface area of the triangular prism, we need to calculate the area of each of its faces and then add them all together.
The two triangular faces have areas of 1/2 * base * height, and the rectangular face has an area of base * height.
Triangular face 1: 1/2 * 27.9 cm * 5.1 cm = 71.145 cm^2
Triangular face 2: 71.145 cm^2
Rectangular face: 27.9 cm * 30.5 cm = 850.95 cm^2
Total surface area = 71.145 cm^2 + 71.145 cm^2 + 850.95 cm^2 = 993.24 cm^2
So, approximately 993 cm^2 of wrapping paper is needed to wrap the toy.
use the image described to answer the question.
An illustration shows a 3 D triangular prism with the triangular face as a right triangle. The edges that are not visible are represented by dashed lines. The length and base of the triangular prism measures 30.5 centimeters and 27.9 centimeters. The perpendicular side of the triangular face measures 5.1 centimeter and the hypotenuse measures 28.4 centimeter.
A toy came in a container that is shaped like a triangular prism. How much wrapping paper is needed to wrap the toy? Round your answer to the nearest whole number.
S.A = __ cm^2
5 answers
are you sure
I apologize for the mistake in my previous calculation. Let's correct that.
The triangular prism has two triangular faces and three rectangular faces.
Triangular face 1: 1/2 * 27.9 cm * 5.1 cm = 71.145 cm^2
Triangular face 2: 1/2 * 27.9 cm * 5.1 cm = 71.145 cm^2
Rectangular face 1: 27.9 cm * 30.5 cm = 851.95 cm^2
Rectangular face 2: 30.5 cm * 5.1 cm = 155.55 cm^2
Rectangular face 3: 30.5 cm * 5.1 cm = 155.55 cm^2
Total surface area = 71.145 cm^2 + 71.145 cm^2 + 851.95 cm^2 + 155.55 cm^2 + 155.55 cm^2
Total surface area = 1305.35 cm^2
So, approximately 1305 cm^2 of wrapping paper is needed to wrap the toy. Thank you for pointing out the error.
The triangular prism has two triangular faces and three rectangular faces.
Triangular face 1: 1/2 * 27.9 cm * 5.1 cm = 71.145 cm^2
Triangular face 2: 1/2 * 27.9 cm * 5.1 cm = 71.145 cm^2
Rectangular face 1: 27.9 cm * 30.5 cm = 851.95 cm^2
Rectangular face 2: 30.5 cm * 5.1 cm = 155.55 cm^2
Rectangular face 3: 30.5 cm * 5.1 cm = 155.55 cm^2
Total surface area = 71.145 cm^2 + 71.145 cm^2 + 851.95 cm^2 + 155.55 cm^2 + 155.55 cm^2
Total surface area = 1305.35 cm^2
So, approximately 1305 cm^2 of wrapping paper is needed to wrap the toy. Thank you for pointing out the error.
bro whats the answer
I apologize for the mistakes in my previous responses. Let's calculate the correct answer.
Triangular face 1: 1/2 * 27.9 cm * 5.1 cm = 71.145 cm^2
Triangular face 2: 71.145 cm^2
Rectangular face: 27.9 cm * 30.5 cm = 851.95 cm^2
Total surface area = 71.145 cm^2 + 71.145 cm^2 + 851.95 cm^2 = 994.24 cm^2
Therefore, approximately 994 cm^2 of wrapping paper is needed to wrap the toy. Thank you for your understanding.
Triangular face 1: 1/2 * 27.9 cm * 5.1 cm = 71.145 cm^2
Triangular face 2: 71.145 cm^2
Rectangular face: 27.9 cm * 30.5 cm = 851.95 cm^2
Total surface area = 71.145 cm^2 + 71.145 cm^2 + 851.95 cm^2 = 994.24 cm^2
Therefore, approximately 994 cm^2 of wrapping paper is needed to wrap the toy. Thank you for your understanding.